The axiom of choice is a common set-theoretic axiom with many equivalents and consequences. This tag is for questions on where we use it in certain proofs, and how things would work without the assumption of this axiom.

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Existence of finite sets of infinite set without using AC

Is it possible to prove that every infinite set $B$ has a subset of cardinality $n$, for every natural $n$, without using AC? I know how to prove this claim by induction. In the induction step I chose ...
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On the contradictory nature of large cardinals & choice-like axioms

Compare these two results: Theorem (Scott): $ZFC+V=L\vdash \nexists~\text{Measurable cardinal}$ Theorem (Kunen): $ZFC+AC\vdash \nexists~\text{Reinhardt cardinal}$ Now compare these two ...
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Examples of Extreme Anti-Choice Axioms

Axiom of Choice has many variants like the followings: There is a choice set for every family of non-empty sets. All sets are well-orderable. Of course in many cases one don't need AC to ...
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Separable Hahn-Banach and the axiom of choice

We had in our functional analysis course a proof for the Hahn-Banach theorem on a separable Banach space which doesn't need, according to our professsor, the axiom of choice. Yesterday I read the ...
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1answer
130 views

Why is the axiom of choice not taught from the start to mathematics undergraduates?

I've recently discovered that the following theorems require the axiom of choice to be proven: every surjective function has a right inverse. a real-valued function that is sequentially continuous ...
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Probabilistic implications of the existence of non-measurable sets

Measure theory and probability theory are deeply connected through the interpretation of subset measures on the sample space as probabilities of events. A major (and somewhat disturbing) result from ...
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ZF and The Cardinality of The Set of Finite Subsets

In a comment on one of my answers, I claimed that the abelian group generated by a set of $S$ generators, each of order two, could take on any infinite cardinality; this is equivalent to saying that, ...
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is there a known set in ZF, such that we can't find a well order on? [duplicate]

is there a known set in ZF, such that we cant find a well order on? since the axiom of choice $(AC)$ and it's negation is consistent with ZF, i wonder if we have a concrete example of a set $A$ that ...
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1answer
43 views

Compact sets closed in Hausdorff spaces without choice?

An elementary proof that compact sets are closed in Hausdorff spaces involves making arbitrary choices based on the Hausdorff property. Is there a way to avoid invoking choice?
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A very quick way to prove a set is measurable. [duplicate]

All examples of non-measurable subset of $\mathbb{R}$ (in the Lebesgue sense) seem to need the axiom of choice in some way or the other. Hence, can we say: The set $A\subseteq \mathbb{R}$ is ...
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555 views

Can someone point out the flaw in my proof of AC?

I have a fake proof of the axiom of countable choice. Obviously it is not correct, but I cannot see my flaw. Forgive me, I am only learning set theory. Let $\{A_n : n \in \mathbb{N}\}$ be a countable ...
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1answer
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Without AC, it is consistent that there is a function with domain $\mathbb{R}$ whose range has cardinality strictly larger than that of $\mathbb{R}$?

I stumbled across this question earlier, and the top comment on the bottom answer asserts two claims: Without the Axiom of Choice, It is consistent that there exists a function with domain ...
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1answer
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A question dealing with cardinals, and axiom of choice.

I am given sets $A$,$B$ such that there exists $f:A\rightarrow B$ s.t. $f$ is onto $B$. I am trying to show that $B\le A$ Let $b\in B$, consider $\{a\in A \mid f(a) = b\}$, assuming axiom of choice, ...
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2answers
39 views

Axiom of Choice: Family of non-empty sets mutually disjoint or not?

I have noticed that the family of non-empty sets referred to in statements of the Axiom of Choice is sometimes required to be mutually disjoint, and sometimes not. Why is that?
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Mean value theorem and the axiom of choice

There's this theorem in Spivak's book of Calculus: Theorem 7 Suppose that $f$ is continuous at $a$, and that $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for ...
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1answer
29 views

AC and Tychonoff theorem

Although I have proof with me that Tynhonoff theorem implies AC. But I have some difficulties with it: 1. Do we define topology on empty set. If not then in proof of Tynhonoff theorem implies AC we ...
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1answer
32 views

How much choice is needed for the transfer principle?

To construct the hyperreals via ultrapower the Boolean prime ideal theorem apparently suffices. However, to prove the transfer principle for the extension $\mathbb{R}\subset{}^\ast\mathbb{R}$ ...
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A countably infinite set of prisoners on death row who believe in the Axiom of Choice are locked in a room. [duplicate]

A countably infinite set of prisoners on death row who believe in the Axiom of Choice and who can process an infinite amount of information in a finite time are locked in a room. The ...
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1answer
35 views

Collection of surjective functions implies axiom of choice

if I have this: (a) If $\left \{ f_i:A_i\rightarrow B_i|i\epsilon I \right \}$ is a collection of surjective functions then $\prod_{i\epsilon I} f_i: \prod_{i\epsilon I} A_i\rightarrow ...
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3answers
651 views

Do we really need Choice to pick socks?

It is said that you need the Axiom of Choice to pick one sock from each of infinitely many pairs, but that you don't need it for shoes, since you can just pick all the left shoes. But Choice is ...
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1answer
74 views

Uncountable series without axiom of choice

Consider a sequence of positive real numbers $(\alpha_i)_{i\in I}$ for some (suppose maybe wellordered for now) set $I$. Using axiom of choice, it is easy to see that $\sum_i \alpha_i$ is infinite if ...
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Differentiability and Monotonic Functions

I just read proof from Royden of theorem: 'Every Monotonic functions are differentiable almost everywhere.' But proof use Vitali Covering Lemma. But Vitali Covering Lemma is based on fact if we assume ...
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2answers
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Countable collection of countable sets and Axiom of choice

Do we need Axiom of choice(or weaker version axiom of countable choice) to say countable Cartesian product of countable sets is nonempty? I think yes. I read somewhere answer no giving argument: each ...
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Infinite prisoners with hats — is choice really needed?

The problem is this (recently asked about here): A countably infinite number of prisoners, each with an unknown and randomly assigned red or blue hat line up single file line. Each prisoner faces ...
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1answer
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Show that, using the axiom of choice, that the cardinality of the sets of all countable subsets of $\mathbb{R}$ have cardinality $2^{\aleph_0}$

Show that, using the axiom of choice, that the cardinality of the sets of all countable subsets of $\mathbb{R}$ have cardinality $2^{\aleph_0}$ and show where it was used the axiom of choice. ...
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Functions that satisfy $f(x+y)=f(x)f(y)$ and $f(1)=e$

My real analysis professor mentioned in passing that there exist functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy all of the following conditions for all $a,b \in \mathbb{R}$: $$f(1)=e$$ ...
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Existence of minimal sub-systems

A topological dynamical system is a topological space $X$ together with a continuous function $f : \ X \to X$. In the following, I will assume that $X$ is compact and Hausdorff (in other words, I work ...
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1answer
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Fixed Point Equivalences of Axiom of Choice

Axiom of Choice has many known equivalences. Also there are many known fixed point theorems (unproved statements) which provide useful information about existence of fixed points for particular ...
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How much conservative ZF+AC and ZF+DC are over ZF?

A logical theory $T_2$ is a (proof theoretic) conservative extension of a theory $T_1$ if the language of $T_2$ extends the language of $T_1$; every theorem of $T_1$ is a theorem of $T_2$; and any ...
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What is the most important (outside of mathematics) results known to require the axiom of choice? [duplicate]

What is the most important (outside of mathematics) results known to require the axiom of choice? I'm skeptical of the utility of the axiom of choice. Of course I realize that it CAN be used, but ...
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1answer
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On conglomerates' axiom of choice(Category theory)

There is a requirement of conglomerate(collection of classes) which demands the following property. Axiom of choice for conglomerates: For each surjection between congomerates $f:X\to Y$, there is an ...
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$x$ is an adherent point of $X$ iff $\exists (a_n)_{n=0}^\infty$ such that $a_n \in X$, which converges to $x$.

Lemma Let $X$ be a subset of $\Bbb R$, and let $x\in\Bbb R$. Then $x$ is an adherent point of $X$ if and only if there exists a sequence $(a_n)_{n=0}^\infty$ consisting entirely of elements in $X$, ...
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What is the “opposite” of the Axiom of Choice?

One might think that, trivially, the "opposite" of AC is $\neg$AC. However, thinking about it differently, I'm not sure this is intuitively the case. AC says that every set has a choice function. ...
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1answer
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Using Zorn's lemma to show a set is measurable.

If $S_j\subset \mathbb{R}^n$ and for all $j\in \mathbb{N}$ the set $S_j$ is measurable, show that $\bigcup S_j$ and $\bigcap S_j$ are measurable. I think I can use Zorn's lemma to show that this is ...
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1answer
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Consistency strength of the “club ultrafilter”

What are the consistency strengths of $$ZF+``\text{The club filter on $\omega_1$ is an ultrafilter}"$$ and $$ZF + DC + ``\text{The club filter on $\omega_1$ is an ultrafilter}"?$$ I know that the ...
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1answer
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Axiom of Choice Equivalent

I'm trying to prove the following statement is equivalent to the Axiom of Choice: "For any set $A$, there exists a function $F$ with dom $F = ⋃A$ and for each $x ∈ ⋃A$, $x ∈ F(x) ∈ A$." (1) The ...
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Does the statement ''PID of dimension $0$ $\Longrightarrow$ field'' actually use Zorn's Lemma?

Everywhere I look seems to blow by the statement that PIDs which are not fields have Krull dimension $1$. This relies on the fact: A PID with Krull dimension $0$ is a field. (*) It seems that the ...
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Does this version of Schröder-Bernstein-Cantor imply choice? [duplicate]

Consider the following statement: $(*)$ For all sets $A$,$B$ and surjections $f\colon A \rightarrow B$, $g\colon B \rightarrow A$ there is a bijection $h\colon A \rightarrow B$ Given choice, this ...
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1answer
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Generic in Boolean-Valued-Models

Let $M$ be a transitive $\in$-interpretation of a extension $T$ of $ZF$ in $ZF$,and let $B$ such that $$T\vdash B\in M\wedge B\text{ is a complete Boolean algebra}$$ Then, using the fact that any set ...
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Bases of complex vector spaces and the axiom of choice

In Zermelo-Fraenkel set theory $ZF$ consider the following statement defined for every field $K$: $B_K$ : Every vector space over $K$ has a basis. It is well-known that $AC \Rightarrow \forall K ...
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1answer
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Proof Explanation: Existence of “fg-chains” Axiom of Choice implies Zorn's Lemma

I would be very glad if someone could explain the validity of a passage of this paper to me. In the proof of Lemma 3.3, "Fundamental Lemma", the notion of "fg-chains" is introduced and later on, we ...
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1answer
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What kind of Choice am I making in this argument?

I have an argument that's supposed to imply Choice, but I'm afraid it may be using some choice. If it does, how much choice? This is the part of the argument that might use some Choice. I marked the ...
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1answer
67 views

Infinite Cardinal Addition Without the Axiom of Choice

In the book 'Introduction to Set Theory' by Hrbacek and Jech, cardinal addition is defined as $$\sum_{i \in I}{\kappa_i}=\left|\bigcup_{i \in I}{A_i} \right|$$ where $|A_i|=k_i$ for all $i \in I$ ...
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Cardinality of the union of two infinite set

Suppose that $A$ and $B$ are two infinite sets and $|A|<|B|$. The question is that how to prove that $|A∪B|=|B|$. The proof is related to the Axiom of Choice.
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Existence of non-trivial linear functional on any vector space

For every vector space $V$ does there exist a linear functional $f$ ( a linear map from $V$ to $F$ the underlying field ) such that for some $ \vec v \in V$ , $f(\vec v) \ne 0$ ? If it does exist , ...
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$\mathbf{Q}$ basis of $\mathbf{R}$.

Could someone give me an explicit basis of $\mathbf{R}$ as a vector space over $\mathbf{Q}$? I no some linearly independent subset, namely $1,e,e^2,\ldots$ but this seems to be a deep result already ...
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1answer
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Does the existence of a recursive sequence that involves an arbitrary choice at every step require the axiom of choice?

Say I want to define a sequence $(x_n)$ recursively, and at each step I make an arbitrary selection for $x_{n+1}$ out of some nonempty pool of acceptable candidates dependent on $x_n$. Does the ...
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1answer
68 views

How big can the continuum be without choice?

I've heard an argument before (although I can't remember where) that the continuum hypothesis is false, since the powerset operation is a something much more 'powerful' than the mere cardinal ...
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Over ZF, does “every Hilbert space have a basis” imply AC?

I know there is a similar result due to Blass [1] that over ZF, "every vector space has a (Hamel) basis" implies AC. Looking around, however, I can't find any results on the question for Hilbert ...
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Existence of a basis in constructive vector spaces

As I was trying to review forgotten knowledge on Vector Spaces in wikipedia, I read that the existence of a basis follows from Zorn lemma, hence equivalently from the axiom of choice. Actually, the ...